If the work done by a battery to move a unit charge through the battery from one terminal to the other is doubled, what is the e
1 answer:
The emf is doubled
Explanation:
The work done by a battery to move a unit charge through the battery from one terminal to the other is equal to the change in electric potential energy of the charge, therefore it can be written as
where
q is the charge
is the potential difference through which the charge has been moved
Since the charge is moving through the battery, this means that the potential difference is equal to the emf of the battery , therefore
(1)
In this problem, we are told that the work done by the battery is doubled, so
From eq.(1), we see that the emf of the battery is proportional to the work done: therefore, since the work doubles, the battery's emf doubles as well.
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Answer:
See the answers below.
Explanation:
The total power of the circuit is equal to the sum of the powers of each lamp.
Now we have a voltage source equal to 240 [V], so by means of the following equation we can find the current circulating in the circuit.
where:
P = power [W]
V = voltage [V]
I = current [amp]
So this is the answer for c) I = 0.67 [amp]
We know that the voltage of each lamp is 240 [V]. Therefore using ohm's law which is equal to the product of resistance by current we can find the voltage of each lamp.
a)
where:
V = voltage [V]
I = current [amp]
R = resistance [ohms]
Therefore we replace this equation in the first to have the current as a function of the resistance and not the voltage.
b)
The effective resistance of a series circuit is equal to the sum of the resistors connected in series.